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pishuonlain [190]
3 years ago
5

A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo

ut 61.0% of the roof. A typical annual rainfall is about 14.0 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water in a full tank of that size weigh (in N and in lbf)?
Engineering
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

volume  = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'\times H

where A' = 61% of A

              = 0.61\times 2667 = 1626.87 sq ft

volume  =  1626.87 \times (\frac{14}{12} ft)

               =1898.015 ft^3

in\ m^3 = \frac{ 1898.015}{35.315} =   53.7457 m^3

in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon

weight = \rho Vg

       = 1000\times 53.74\times 9.8

             =526652 N

In\ lbf =  \frac{526652}{4.448} = 118396.08 lbf

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Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur
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Answer:

Explanation:

Answer:

Explanation:

Answer:  

Explanation:  

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.  

(a). here we are asked to determine the Temperature and Pressure.  

Given that the properties of Air;  

ha = 230.02 KJ/Kg  

Ta = 230 K  

Pra = 0.5477  

From the energy balance equation for a diffuser;  

ha + Va²/2 = h₁ + V₁²/2  

h₁ = ha + Va²/2 (where V₁²/2 = 0)  

h₁ = 230.02 + 220²/2 ˣ 1/10³  

h₁ = 254.22 KJ/Kg  

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg  

from this we have;  

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]  

Pr₁ = 0.77759  

therefore T₁ = 254.15K  

P₁ = (Pr₁/Pra)Pa  

= 0.77759/0.5477 ˣ 26  

P₁ = 36.91 kPa  

now we calculate Pr₂  

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349  

⇒ now we obtain properties of air at  

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg  

calculating the enthalpy of air at state 2  

ηc = h₁ - h₂ / h₁ - h₂  

0.85 = 254.22 - 505.387 / 254.22 - h₂  

h₂ = 549.71 KJ/Kg  

to obtain the properties of air at h₂ = 549.71 KJ/Kg  

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913  

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at  

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁  

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄  

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃       i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄  + 0  = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

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Who can use NIST resources?
sukhopar [10]

Answer:

Federal agencies

Explanation:

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As part of this mission, NIST scientists and engineers continually refine the science of measurement (metrology) by creating precise engineering and manufacturing required for most current technological advances. They are also directly involved in the development and testing of standards made by the private sector and government agencies. The NIST was originally called the National Bureau of Standards (NBS), a name it had from 1901 to 1988. The progress and technological innovation of the United States depends on the abilities of the NIST, especially if we talk about four areas: biotechnology , nanotechnology, information technologies and advanced manufacturing.

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